I am having so much trouble on this! "the length of a rectangle is 6 more than twice the width. Find the dimensions of the rectangle if the area is 56 square inches."it is a math question where you...

I am having so much trouble on this! "the length of a rectangle is 6 more than twice the width. Find the dimensions of the rectangle if the area is 56 square inches."it is a math question where you have to solve using factoring methods

Here's another way to solve it. Since you know the area is 56, and that the area is l x w, you can easily come up with a list of all the possible combinations that involve at least one side being a whole number.

56 x 1

28 x 2

18.66 x 3

14 x 4

11.2 x 5

9.33 x 6

7 x 8

The next would be 8 x 7, so you may as well stop.

Now see if any of these fit the formula of one number being equal to 2(the number) + 6.

56 x 1 -- 2 x 1 = 2, which is far more than 6 away from 56.

The same can be said of 28 x 2.

18.66 x 3 is closer. However, 3 x 2 = 6, and 18.66 is more than 6 away from that.

14 x 4 is perfect!! 4 x 2 = 8, and 8 + 6 = 14.

Just to be safe, check the next to see if it's less than 6 away -- yes! 5 x 2 = 10, which is less than 6 away from 11.2.

The first thing you need to do is translate these words into the numbers and operations you need in an equation. The words tell you that the description is all equal to 56 squared. So, you already have one side of the equation. What do you need to do for the other side? "6 more" translates into "6 +." If we call the width "w," then we can say that "twice the width" equals "2w." How do we get an area? By multiplying length times width. We know that "w" is the width, and we know that the length is "2w + 6." Now, you have two sides for your equation, an equation in which you have to solve for "w."

Here is how this looks:

(w)(2w + 6) = 56 squared.

Now you can solve the equation algebraically. If you carry out the operation on the left,you might be able to factor into squares and then take the square root of each side and solve for "w" and then solve for "l" by substitution.